Divide the following complex numbers. $ \dfrac{-1+13i}{2-i}$
We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${2+i}$ $ \dfrac{-1+13i}{2-i} = \dfrac{-1+13i}{2-i} \cdot \dfrac{{2+i}}{{2+i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(-1+13i) \cdot (2+i)} {(2-i) \cdot (2+i)} = \dfrac{(-1+13i) \cdot (2+i)} {2^2 - (-1i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(-1+13i) \cdot (2+i)} {(2)^2 - (-1i)^2} = $ $ \dfrac{(-1+13i) \cdot (2+i)} {4 + 1} = $ $ \dfrac{(-1+13i) \cdot (2+i)} {5} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-1+13i}) \cdot ({2+i})} {5} = $ $ \dfrac{{-1} \cdot {2} + {13} \cdot {2 i} + {-1} \cdot {1 i} + {13} \cdot {1 i^2}} {5} $ Evaluate each product of two numbers. $ \dfrac{-2 + 26i - 1i + 13 i^2} {5} $ Finally, simplify the fraction. $ \dfrac{-2 + 26i - 1i - 13} {5} = \dfrac{-15 + 25i} {5} = -3+5i $